\beginsection 19.2

Prove that each of the following functions is uniformly continuous on the
indicated set by directly verifying the $\epsilon$-$\delta$ property in
Definition 19.1.
\medskip
To verify the $\epsilon$-$\delta$ property, start with the expression
$|x-y|<\delta$ and then find a way to transform it into the expression
$|f(x)-f(y)|<\epsilon$.

\medskip
(a) $f(x)=3x+11$ on $R$.
\medskip
We have
$$|f(x)-f(y)|=|3x+11-3y-11|=3|x-y|$$
Starting with
$$|x-y|<\delta$$
we can make the substitution
$$|f(x)-f(y)|<3\delta$$
Then for $\epsilon=3\delta$ we have
$$|f(x)-f(y)|<\epsilon$$

\medskip
(b) $f(x)=x^2$ on $[0,3]$
\medskip
We have
$$|f(x)-f(y)|=|x^2-y^2|=|x+y|\cdot|x-y|$$
and
$$|f(x)-f(y)|\le6|x-y|$$
on the interval $[0,3]$.
Starting with
$$|x-y|<\delta$$
we can make the substitution
$$|f(x)-f(y)|<6\delta$$
Then for $\epsilon=6\delta$ we have
$$|f(x)-f(y)|<\epsilon$$

\medskip
(c) $f(x)=1/x$ on $[{1\over2},\infty)$
\medskip
We have
$$|f(x)-f(y)|=\left|{1\over x}-{1\over y}\right|
=\left|{y-x}\over xy\right|
={|x-y|\over|xy|}$$
and
$$|f(x)-f(y)|\le4|x-y|$$
on the interval $[{1\over2},\infty)$.
Starting with
$$|x-y|<\delta$$
we can make the substitution
$$|f(x)-f(y)|<4\delta$$
Then for $\epsilon=4\delta$ we have
$$|f(x)-f(y)|<\epsilon$$
